Question

If x^2+y^2=29xy then prove that 2log(x-y)=3log3+logx+logy.

Answer 1

Subtract 2xy from lhs and rhs
X^2 + y^2 - 2xy = 27xy
(x-y)^2 = 27xy
Taking log both the sides
Log (x-y)^2 = log 27xy which is further = to log27 + logx +logy
By the properties of power
log(x-y)^2 = 2 × log (x-y) and
log27=3× log 3
HENCE PROVED

Answer 2

Solution :-

x² + y² = 29xy

Subtracting 2xy on both sides

x² + y² - 2xy = 29xy - 2xy

x² + y² - 2xy = 27xy

(x - y)² = 27xy

[ Because (x - y)² = x² + y² - 2xy ]

Taking log on both sides

log(x - y)² = log27xy

log(x - y)² = log27 + logx + logy

$\boxed{ \bf \because logab = loga + logb}$

log(x - y)² = log3³ + logx + logy

2log(x - y) = 3log3 + logx + logy

$\boxed{ \bf \because loga^m = m.loga}$

Hence proved

Laws of logarithms used :-

$\\ \sf \longrightarrow logab = loga + logb \\ \\ \\ \sf \longrightarrow loga^m= m.loga$